Aptitude · Compound Interest · Model 4

The interest earned in one particular year

Q: Give the general formula for the nth year's interest.

P(1 + R/100)^(n−1) × R/100 . Grow the principal for n−1 years to get the balance at the start of year n, then take one year’s interest on it.

“Find the interest for the third year” is not the same as “find the interest for three years”. One is a single year’s earnings on a grown balance; the other is everything so far. Reading which is wanted is most of the model.

Pick a year and see which balance it charges on →

Interest in year n = P(1 + R/100)ⁿ⁻¹ × R/100 — one year’s interest on the balance standing at the start of that year.

01 The idea

In the year, not for the years

On ₹10,000 at 10%, the interest earned during the third year is ₹1,210. The interest earned over three years is ₹3,310. Both are correct answers to different questions, and examiners rely on students answering the second when the first was asked.

The single-year figure is easy once you see what base it uses. Year three charges 10% on whatever was standing at the start of year three — that is, after two years of growth. Ten thousand grown for two years is ₹12,100, and 10% of that is ₹1,210.

So the general form writes itself: grow the principal for n − 1 years, then take one year’s interest on the result. The exponent is n − 1 and not n, which is the detail that decides the answer.

There is a pattern worth noticing. On ₹10,000 at 10% the yearly interests run 1,000, 1,100, 1,210, 1,331 — each one is the previous multiplied by 1.1. The yearly interests form a geometric sequence with the same ratio as the balances, which makes questions about consecutive years quick.

Year n charges interest on the balance after n − 1 years of growth. The exponent is n − 1, and getting that right is the model.

Interest in the nth yearThe interest earned during that single year alone: P(1 + R/100)ⁿ⁻¹ × R/100.

Interest for n yearsThe total accumulated: P(1 + R/100)ⁿ − P. A different and larger quantity.

Geometric progression of interestsSuccessive yearly interests are in the ratio (1 + R/100). So each year’s interest is the last one grown by the same rate.

02 Worked example

₹10,000 at 10% — the third year alone

The distinction, made concrete. ₹10,000 is invested at 10% per annum compound interest. Find the interest earned during the third year, and contrast it with the interest earned over three years.

Ask what year three charges onInterest for year three is charged on the balance at the start of year three, which is after two full years of growth.base for year 3 = 10,000 × (1.1)²

Grow the principal for two yearsTwo years, so the exponent is two — one less than the year number.10,000 × 1.21 = ₹12,100

Take one year of interest on thatA single 10%, not compounded again. This is the answer to the question asked.10% of 12,100 = ₹1,210

Now the total over three yearsA different quantity: the amount after three years, less the principal.10,000 × 1.331 − 10,000 = ₹3,310

Check the pieces add upThe three yearly interests should total the three-year interest.1,000 + 1,100 + 1,210 = ₹3,310 ✓

The check in the last step is the best way to keep the two ideas apart: the single-year figures are the parts and the multi-year figure is their sum. Note also that 1,000, 1,100 and 1,210 each exceed the last by a factor of 1.1 — so if you know any one year’s interest, you can get the neighbouring years by multiplying or dividing by the same multiplier, with no powers at all.

03 The method

The formula, and the shortcuts the pattern gives you

One formula, plus two consequences of the geometric pattern that turn several question types into one multiplication.

Interest in year n = P(1 + R/100)ⁿ⁻¹ × R/100. Compare interest for n years = P(1 + R/100)ⁿ − P.

Consecutive years are one multiplication apart. Interest in year n+1 is interest in year n times (1 + R/100). So if the second year’s interest is ₹1,100 at 10%, the third is ₹1,210 with no further work. And the difference between consecutive years’ interests is itself just R% of the earlier year’s interest.

Year	Base it charges on	Interest at 10% on ₹10,000
1	`P`	`1,000`
2	`P(1.1)`	`1,100`
3	`P(1.1)²`	`1,210`
4	`P(1.1)³`	`1,331`
n	`P(1.1)^(n−1)`	`P(1.1)^(n−1) × 0.1`
Total, 3 years	—	`3,310 = the sum`
Using exponent n	`wrong`	gives year n+1's interest

05 Cheat sheet

Model 4 on one page

One formula and three consequences of the geometric pattern.

Want	Route	On ₹10,000 at 10%
Interest in year n	`P(1+R/100)^(n−1) × R/100`	yr 3 → 1,210
Interest for n years	`P(1+R/100)^n − P`	3 yr → 3,310
Next year's interest	`× (1 + R/100)`	1,100 → 1,210
Previous year's interest	`÷ (1 + R/100)`	1,210 → 1,100
Difference between consecutive	`R% of the earlier interest`	10% of 1,100 = 110
Year 1 interest	`same as simple interest`	1,000
Using exponent n	`off by one year`	gives 1,331, not 1,210

The exponent is n minus 1Year n charges on the balance after n−1 years of growth. Using n gives the following year's interest, which will be among the options.

Yearly interests form a GPEach is the previous one times (1 + R/100). So neighbouring years are one multiplication apart, with no powers needed.

The parts sum to the wholeThe individual yearly interests add up to the multi-year compound interest. Use that as a check whenever you are unsure which quantity you computed.

06 Where & why

Where Model 4 shows up

A precision model: it tests whether you read the question, and whether you know which balance a percentage applies to.

Bank PO Mains · SSC CGL

“Interest for the third year”

Set specifically because the total-for-three-years answer is the natural mistake and sits in the options.

Difference between two years

“How much more in year 3 than year 2?”

Just R% of year 2's interest. Recognising that avoids computing two powers and subtracting.

Reverse questions

Given one year's interest, find the principal

Divide out the multiplier and the rate. These are common and are much easier than they look once the exponent is right.

Instalments and amortisation

Interest in a particular period

Real loan schedules quote exactly this figure for each period, so the model reflects how repayment tables are actually built.

Before calculating, underline whether the question says “in the nth year” or “in n years”. Those two phrases are the entire difficulty of this model, and everything else is one multiplication.

07 Interview questions

What gets asked

Nine, and the first two are the distinction the model is built on.

What is the interest earned in the third year on ₹10,000 at 10% compound interest?

₹1,210. Year three charges on the balance after two years of growth, which is 10,000 × 1.21 = ₹12,100, and 10% of that is ₹1,210. The exponent is two, not three.

How is that different from the interest for three years?

The interest for three years is the total accumulated: 10,000 × 1.331 − 10,000 = ₹3,310. The year-three figure of ₹1,210 is just one of the three yearly pieces, and the three pieces — 1,000, 1,100 and 1,210 — add to ₹3,310.

Give the general formula for the nth year's interest.

P(1 + R/100)^(n−1) × R/100. Grow the principal for n−1 years to get the balance at the start of year n, then take one year’s interest on it.

Why is the exponent n minus 1 rather than n?

Because interest for a year is charged on the balance at the start of that year, and at the start of year n only n−1 years of growth have happened. Using n computes year n+1’s interest instead, which is a plausible wrong answer.

What pattern do the successive yearly interests follow?

A geometric progression with ratio (1 + R/100). On ₹10,000 at 10% they are 1,000, 1,100, 1,210, 1,331 — each 1.1 times the last. So if you know one year’s interest you get its neighbours by multiplying or dividing by 1.1.

The interest in the second year is ₹1,100 at 10%. Find the third year's interest and the principal.

Third year: ₹1,210, just 1,100 × 1.1. The principal: year two’s interest is P × 1.1 × 0.1, so P = 1100/(0.11) = ₹10,000.

How much more interest does year 3 earn than year 2?

Exactly R% of year 2’s interest. On ₹10,000 at 10% that is 10% of ₹1,100 = ₹110, which checks against 1,210 − 1,100. This avoids computing two powers and subtracting them.

Is the first year's interest under compounding the same as simple interest?

Hard

₹10,000 is invested at 10% per annum compound interest for 4 years. (a) List the interest earned in each of the four years. (b) Verify that they sum to the four-year compound interest. (c) Show algebraically that for any P, R and n, the sum of the individual yearly interests equals P(1 + R/100)^n − P, and identify what kind of series you have summed.

Follow-up

Part (c) makes the check in part (b) into a proof: the yearly interests form a geometric series, and summing it telescopes to exactly the compound interest formula. That is a satisfying connection, and it explains why the parts-and-whole check always works rather than merely happening to.

Show the hint

For (c), write the year-k interest as P(1+r)^(k−1)·r and sum from k = 1 to n using the geometric series formula.